A photoelectron produced by ionization in a photoelectron spectrometer is ejected with a velocity of 861 km s^-1. Calculate the de Broglie wavelength of the electron. A) 8.45 nm B) 8.45 Å C) 8.45 pm D) 0.845 $\mu$m

Question

Quizplus · Accepted Answer

The de Broglie wavelength ($\lambda$) of a particle is given by the equation $\lambda = \frac{h}{mv}$, where $h$ is Planck's constant ($6.626 \times 10^{-34} m^2 kg / s$), $m$ is the mass of the electron ($9.109 \times 10^{-31} kg$), and $v$ is the velocity of the electron ($861 \times 10^3 m/s$). Substituting these values into the equation gives a wavelength in meters, which when converted to angstroms ($1 \text{Å} = 10^{-10} m$), results in approximately 8.45 Å.

A Photoelectron Produced by Ionization in a Photoelectron Spectrometer Is μ\muμ

A Photoelectron Produced by Ionization in a Photoelectron Spectrometer Is $\mu$